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AQA A-level Physics revision notes

Medical physics (A-level only)

Section 3.10
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
19 specification points
Optional · choose 1 of 5

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.10

Checked against AQA 7408 section 3.10. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

In the exam: Data and formulae booklet provided · calculator allowed in every paper

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3.10.1.1

Physics of vision

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The eye is an optical refracting system. The cornea provides most refraction, while the lens changes power during accommodation so that a real, inverted image forms on the retina.
  • Ray diagrams should show refraction at the eye and convergence onto the retinal surface.
  • Cones work best in bright light, give colour vision and provide high spatial resolution because there is little convergence of their nerve pathways.
  • Rods are more sensitive at low light levels and many feed one nerve pathway, increasing sensitivity but reducing spatial resolution; rods do not distinguish colour.
  • The eye's sensitivity also depends on wavelength, so its spectral response changes with the active receptor cells and lighting conditions.
Parallel rays bend first at the cornea and again through the separate biconvex lens before converging on the retina.
Worked example

Explain why cone-rich vision gives greater spatial resolution than rod-rich vision.

  1. 1.Cone pathways show little convergence, so signals from nearby receptors remain distinguishable.
  2. 2.Many rods share a nerve pathway, so separate nearby stimuli may produce the same combined signal.
  3. 3.The reduced convergence of cone pathways therefore preserves more positional detail.

Answer: Cones give greater spatial resolution because their low pathway convergence keeps signals from neighbouring retinal positions separate.

Common mistakes

  • Don't state that the lens provides all of the eye's refraction and omit the cornea.
  • Don't describe rods as producing detailed colour vision rather than sensitive low-light vision.
  • Don't claim that convergence of many receptors onto one nerve pathway improves spatial resolution.

Exam tip

A compare question should link receptor wiring to performance: greater convergence increases sensitivity but reduces spatial resolution.

Tier 1 · Easy

ORIGINAL

State the type of retinal receptor that is mainly responsible for vision at very low light intensity and give one property of the resulting image.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Explain why two faint points of light that fall on nearby rod cells may be seen as one point even though both points are detected.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A coloured grid is viewed first in bright light and then after the illumination is greatly reduced. Explain the changes expected in colour, detail and sensitivity, referring to the retinal receptors and their nerve connections.

[5 marks]

Total for this question: 5

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3.10.1.2

Defects of vision and their correction using lenses

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Lens power is P=1/fP=1/f, with ff in metres and power in dioptres; a converging lens has positive power and a diverging lens negative power. Using the real-is-positive convention, 1/f=1/u+1/v1/f=1/u+1/v and m=v/um=v/u, so a virtual image distance is negative.
  • Myopia makes distant objects focus in front of the retina and is corrected by a diverging lens.
  • Hypermetropia prevents nearby objects focusing on the retina and is corrected by a converging lens.
  • Astigmatism gives different focusing in different planes and is corrected using a cylindrical component; a prescription states its cylindrical power and an axis from 00^\circ to 180180^\circ.
  • Ray diagrams must show how the correcting lens changes convergence.
Worked example

A myopic eye has a far point 0.80m0.80\,\text{m} from the eye. Calculate the power of a correcting lens that makes a distant object appear at the far point.

  1. 1.For a distant object, u=u=\infty, and the required virtual image has v=0.80mv=-0.80\,\text{m}.
  2. 2.Use 1/f=1/u+1/v=01/0.80=1.25m11/f=1/u+1/v=0-1/0.80=-1.25\,\text{m}^{-1}.
  3. 3.Since P=1/fP=1/f, the correcting-lens power is 1.25D-1.25\,\text{D}.

Answer: A diverging lens of power 1.25D-1.25\,\text{D} is required.

Common mistakes

  • Don't use a positive image distance for the virtual image at the myopic eye's far point.
  • Don't calculate P=1/fP=1/f with focal length in centimetres rather than metres.
  • Don't correct hypermetropia with a diverging lens instead of a converging lens.

Exam tip

State the sign and lens type with a calculated prescription; a negative dioptre value must be identified as diverging.

Tier 1 · Easy

ORIGINAL

State the type of spectacle lens used to correct myopia and state the sign of its power.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A spectacle lens of power +6.25D+6.25\,\text{D} corrects a hypermetropic eye. An object is 0.400m0.400\,\text{m} from the lens. Using the real-is-positive convention, calculate the image distance and the magnitude of the magnification.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A hypermetropic eye has a near point 0.80m0.80\,\text{m} from a spectacle lens. Determine the lens power needed so that an object 0.25m0.25\,\text{m} from the lens forms a virtual image at the near point. Use the real-is-positive convention.

[5 marks]

Total for this question: 5

3.10.2.1

Ear as a sound detection system

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The pinna and ear canal direct pressure variations in air to the tympanic membrane, which vibrates at the sound frequency.
  • The ossicles transmit this mechanical vibration through the middle ear and act as a lever system.
  • Because the oval window has a smaller area than the tympanic membrane, the transmitted force produces a larger pressure in the inner-ear fluid.
  • Pressure waves in the inner-ear fluid stimulate receptors, which produce electrical impulses in the auditory nerve.
  • A complete explanation follows the energy transfer in order: air pressure variation, membrane vibration, ossicle motion, fluid pressure wave, receptor stimulation and electrical nerve signal.
Worked example

Explain why the ossicles and oval window help transfer sound from air into inner-ear fluid.

  1. 1.The ossicles act as levers and transmit force from the tympanic membrane.
  2. 2.The oval window has a smaller area than the tympanic membrane.
  3. 3.For the transmitted force, the smaller area produces a greater pressure in the inner-ear fluid.

Answer: The lever action and area reduction increase fluid pressure, improving transmission into the inner ear.

Common mistakes

  • Don't send sound directly from the ear canal into the inner ear and omit the tympanic membrane and ossicles.
  • Don't say that the larger oval-window area increases pressure even though pressure is force divided by area.
  • Don't describe the auditory nerve signal as a sound wave rather than an electrical signal.

Exam tip

For a transmission question, give the full ordered chain and name the change from mechanical motion to electrical signalling.

Tier 1 · Easy

ORIGINAL

Name the membrane that vibrates when sound reaches the end of the ear canal and name the three-bone system moved by it.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Explain two features of the middle ear that increase the pressure delivered to the inner-ear fluid.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A sound produces pressure amplitude 0.012Pa0.012\,\text{Pa} on a tympanic membrane of area 5.5×105m25.5\times10^{-5}\,\text{m}^2. The ossicles increase the force by a factor of 1.41.4 and act on an oval window of area 3.2×106m23.2\times10^{-6}\,\text{m}^2. Calculate the pressure amplitude delivered to the inner-ear fluid.

[5 marks]

Total for this question: 5

3.10.2.2

Sensitivity and frequency response

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Sound intensity is power transferred per unit area, measured in W m2\text{W m}^{-2}; an isotropic point source gives I=P/(4πr2)I=P/(4\pi r^2). Intensity level is L=10log10(I/I0)L=10\log_{10}(I/I_0) in decibels, where I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.
  • The logarithmic scale reflects perception and compresses a very wide intensity range.
  • Equal-loudness curves show the level required at each frequency for the same perceived loudness; lower curves mean greater sensitivity, normally around a few kilohertz.
  • A relative level is ΔL=10log10(I2/I1)\Delta L=10\log_{10}(I_2/I_1).
  • The dBA scale applies frequency weighting to approximate human response, whereas an unweighted dB value represents the physical intensity ratio alone.
Equal-loudness curves showing greatest hearing sensitivity at a few kilohertz.
Worked example

Calculate the intensity level for I=1.0×106W m2I=1.0\times10^{-6}\,\text{W m}^{-2}.

  1. 1.Use L=10log10(I/I0)L=10\log_{10}(I/I_0) with I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.
  2. 2.The ratio is I/I0=106I/I_0=10^6.
  3. 3.Therefore L=10log10(106)=60dBL=10\log_{10}(10^6)=60\,\text{dB}.

Answer: The intensity level is 60dB60\,\text{dB}.

Common mistakes

  • Don't use 20log1020\log_{10} for an intensity ratio instead of the required factor 1010.
  • Don't read the highest point of an equal-loudness curve as the frequency of greatest sensitivity.
  • Don't treat dBA frequency weighting as identical to an unweighted dB measurement.

Exam tip

When comparing two sounds, use the intensity ratio directly in ΔL=10log10(I2/I1)\Delta L=10\log_{10}(I_2/I_1).

Tier 1 · Easy

ORIGINAL

Calculate the intensity level of a sound with intensity 1.0×108W m21.0\times10^{-8}\,\text{W m}^{-2}. Use I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A sound detector reads an intensity level of 83dB83\,\text{dB}. Its collecting area is 2.0×103m22.0\times10^{-3}\,\text{m}^2. Calculate the sound intensity at the detector and the sound power incident on it. Use I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An isotropic sound source has power 0.80W0.80\,\text{W}. Determine the distance at which its intensity level is 85dB85\,\text{dB}. Then explain why a 12kHz12\,\text{kHz} tone at this level may sound quieter than a 3kHz3\,\text{kHz} tone at the same level. Use I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.

[5 marks]

Total for this question: 5

3.10.2.3

Defects of hearing

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Hearing loss means that a greater sound intensity level is needed to produce the same perceived loudness.
  • On an equal-loudness or threshold graph, the affected curve shifts upward over the frequencies where sensitivity has fallen.
  • The hearing loss at a stated frequency is the vertical difference in decibels between normal and impaired curves, not the horizontal frequency separation.
  • Deterioration with age commonly produces greater loss at high frequencies, while injury from prolonged excessive noise may produce a pronounced loss over a narrower frequency range, often near 4kHz4\,\text{kHz}.
  • The required account concerns the change in sensitivity and its representation on equal-loudness curves; unsupported detail about physiological damage is outside this specification point.
Worked example

At 4.0kHz4.0\,\text{kHz}, a normal threshold is 5dB5\,\text{dB} and an impaired threshold is 35dB35\,\text{dB}. Determine the hearing loss.

  1. 1.Read both levels at the same frequency.
  2. 2.Find the vertical difference: 355=30dB35-5=30\,\text{dB}.

Answer: The hearing loss at 4.0kHz4.0\,\text{kHz} is 30dB30\,\text{dB}.

Common mistakes

  • Don't measure a horizontal frequency difference instead of the vertical dB separation between curves.
  • Don't shift an impaired threshold curve downward even though more intensity is required for detection.
  • Don't give detailed ear physiology when the question asks for the change shown on equal-loudness curves.

Exam tip

State both the affected frequency range and the upward change in required intensity level when interpreting hearing loss.

Tier 1 · Easy

ORIGINAL

State the frequency range in which age-related hearing deterioration is usually greatest.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

At 4.0kHz4.0\,\text{kHz}, a normal threshold is 8dB8\,\text{dB} and an impaired threshold is 38dB38\,\text{dB}. Calculate the hearing loss and the factor by which the threshold intensity has increased.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Listener A has hearing losses of 6dB6\,\text{dB} at 0.5kHz0.5\,\text{kHz}, 15dB15\,\text{dB} at 4kHz4\,\text{kHz} and 44dB44\,\text{dB} at 10kHz10\,\text{kHz}. Listener B has losses of 9dB9\,\text{dB}, 48dB48\,\text{dB} and 17dB17\,\text{dB} at the same frequencies. Deduce the likely cause for each pattern and explain how each pattern changes an equal-loudness curve and perceived sound.

[5 marks]

Total for this question: 5

3.10.3.1

Simple ECG machines and the normal ECG waveform

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Skin electrodes measure small potential differences produced by electrical activity in the heart. Conductive gel, prepared skin and secure contacts reduce resistance and movement artefacts; a high-gain, low-noise differential amplifier and shielded leads are needed because signals are of order millivolts.
  • On a normal waveform, the P wave represents atrial depolarisation, the QRS complex ventricular depolarisation and the T wave ventricular repolarisation.
  • Heart period is measured between equivalent points, usually successive R peaks, and rate=60/T\text{rate}=60/T when TT is in seconds.
  • Measuring across several cycles reduces percentage uncertainty.
  • An ECG records electrical potential difference, not blood pressure, blood flow or the heart's mechanical force.
Normal ECG waveform showing P, QRS and T features and an R-to-R period.
Worked example

Six R-to-R intervals occupy 4.80s4.80\,\text{s} on an ECG trace. Calculate the heart rate.

  1. 1.Find the mean period: T=4.80/6=0.800sT=4.80/6=0.800\,\text{s}.
  2. 2.Use rate=60/T=60/0.800\text{rate}=60/T=60/0.800.

Answer: The heart rate is 75.0beats min175.0\,\text{beats min}^{-1}.

Common mistakes

  • Don't count six R peaks as six complete intervals when the trace contains only five gaps between them.
  • Don't identify the QRS complex as ventricular repolarisation instead of ventricular depolarisation.
  • Don't describe the vertical ECG signal as blood pressure rather than potential difference.

Exam tip

Measure several R-to-R intervals and divide by the number of intervals before converting period to beats per minute.

Tier 1 · Easy

ORIGINAL

Identify the electrical events represented by the QRS complex and the T wave in a normal ECG.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An ECG is recorded on paper moving at 25mm s125\,\text{mm s}^{-1}. The horizontal separation from the start of the P wave to the start of the QRS complex is 4.8mm4.8\,\text{mm}. Calculate the PR interval and determine whether it lies within the supplied normal range 0.120.12 to 0.20s0.20\,\text{s}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An ECG trace contains slow baseline changes when the patient moves and a regular interference signal from nearby mains equipment. Explain how electrode attachment, the leads and the amplifier should be arranged to obtain a clearer normal waveform.

[5 marks]

Total for this question: 5

3.10.4.1

Ultrasound imaging

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Acoustic impedance is Z=ρcZ=\rho c. At a boundary, the reflected intensity fraction is Ir/Ii=((Z2Z1)/(Z2+Z1))2I_r/I_i=((Z_2-Z_1)/(Z_2+Z_1))^2; a larger mismatch gives a stronger echo and less transmission.
  • Coupling gel removes air between transducer and skin, reducing mismatch. A piezoelectric crystal converts a short alternating potential difference into an ultrasound pulse and converts returning deformation into a detected potential difference.
  • Echo depth is d=ct/2d=ct/2 because the pulse travels out and back. A new pulse is sent only after the deepest echo returns, so the maximum pulse-repetition frequency is set by the two-way travel time.
  • Higher frequency improves resolution but increases attenuation. An A-scan displays echo amplitude against time or depth; a B-scan maps echo brightness and position into a two-dimensional image.
  • Ultrasound is non-ionising and real-time, but air and bone boundaries limit access and resolution.
At normal incidence, part of the ultrasound pulse is transmitted and the echo retraces the outward path to the transducer (drawn offset for clarity).
Worked example

An echo returns 78μs78\,\mu\text{s} after emission. Calculate the boundary depth for c=1540m s1c=1540\,\text{m s}^{-1}.

  1. 1.Convert the time: t=78×106st=78\times10^{-6}\,\text{s}.
  2. 2.Use d=ct/2d=ct/2 because the measured time includes both journeys.
  3. 3.Calculate d=1540(78×106)/2=6.01×102md=1540(78\times10^{-6})/2=6.01\times10^{-2}\,\text{m}.

Answer: The boundary is approximately 6.0cm6.0\,\text{cm} deep.

Common mistakes

  • Don't use d=ctd=ct and forget that the echo time covers the outward and return paths.
  • Don't say that coupling gel increases the air gap instead of removing the impedance mismatch with air.
  • Don't call an A-scan a two-dimensional brightness image rather than an amplitude trace.

Exam tip

For an echo calculation, state the factor of two explicitly; for a comparison, balance non-ionising real-time imaging against attenuation and resolution.

Tier 1 · Easy

ORIGINAL

A clinician needs to monitor fetal movement repeatedly during pregnancy and view the movement as it happens. Identify a suitable imaging modality and explain why it is appropriate in terms of image timing and radiation safety.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Ultrasound passes normally from tissue A, where ρ=1000kg m3\rho=1000\,\text{kg m}^{-3} and c=1500m s1c=1500\,\text{m s}^{-1}, into tissue B, where ρ=1060kg m3\rho=1060\,\text{kg m}^{-3} and c=1540m s1c=1540\,\text{m s}^{-1}. Calculate the percentage of incident intensity reflected at the boundary.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A 4.0MHz4.0\,\text{MHz} ultrasound pulse travels through tissue at 1540m s11540\,\text{m s}^{-1} and returns to the transducer 130μs130\,\mu\text{s} after transmission. Calculate the reflector depth and estimate the scan resolution as one wavelength. Explain why coupling gel is placed between the transducer and skin.

[6 marks]

Total for this question: 6

3.10.4.2

Fibre optics and endoscopy

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An optical fibre has a higher-refractive-index core surrounded by lower-index cladding. A ray travelling in the core undergoes total internal reflection when it meets the boundary from higher to lower index at an angle greater than the critical angle.
  • Cladding also keeps light within each fibre and reduces cross-talk.
  • In a flexible endoscope, a non-coherent bundle carries illumination because fibre order need not be preserved.
  • A coherent bundle keeps fibres in the same relative positions at both ends, so each fibre transfers one part of the image.
  • Flexible bundles allow internal imaging through a small opening and can also guide treatment light, reducing the need for more invasive access.
Light guided through a fibre by total internal reflection at the core-cladding boundary.
Worked example

Explain why an endoscope needs a coherent fibre bundle for imaging but may use a non-coherent bundle for illumination.

  1. 1.An image requires each fibre's position at the entrance to match its position at the exit.
  2. 2.A coherent bundle preserves that relative fibre order and therefore the spatial pattern.
  3. 3.Illumination only requires light delivery, so preserving fibre order is unnecessary.

Answer: The coherent bundle transfers the image pattern; the non-coherent bundle can supply illumination.

Common mistakes

  • Don't state only that light reflects, without giving both total-internal-reflection conditions.
  • Don't use a non-coherent bundle to transmit an image even though fibre positions are scrambled.
  • Don't place the cladding at a higher refractive index than the core.

Exam tip

A full total-internal-reflection statement must name travel from higher to lower refractive index and incidence above the critical angle.

Tier 1 · Easy

ORIGINAL

Name the endoscope fibre bundle that carries an image and the bundle that carries illumination into the body.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A fibre has core refractive index 1.621.62 and cladding refractive index 1.501.50. Calculate the critical angle at the core-cladding boundary and determine whether a ray incident there at 72.072.0^\circ undergoes total internal reflection.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A damaged endoscope has an image bundle whose fibre positions are rearranged between its two ends, and some cladding has been removed. Explain the effects on the observed image and why the illumination bundle can still work when its fibre order is random.

[5 marks]

Total for this question: 5

3.10.4.3

Magnetic resonance (MR) scanner

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An MR scanner uses the strong field of a superconducting magnet to align hydrogen nuclei, or protons, with spins parallel; the spinning nuclei precess about the magnetic field lines. Gradient coils vary the field with position so successive small regions of a patient cross-section can be selected and located.
  • A short radio-frequency pulse excites protons in a selected region and changes their spin state. As the protons de-excite after the pulse, they emit radio-frequency signals.
  • The emitted radio-frequency signal has the same frequency as the exciting pulse. Receiver coils detect these signals and a computer processes their position and strength to construct a visual cross-sectional image.
  • MR uses non-ionising radio waves and magnetic fields.
  • Production of the fields and detailed relaxation times are not required.
Worked example

Outline how an MR scanner obtains a cross-sectional image after the patient is placed in the main magnetic field.

  1. 1.Gradient fields select and locate successive small regions of the cross-section.
  2. 2.Short RF pulses excite aligned, precessing hydrogen nuclei in each selected region.
  3. 3.The nuclei emit RF signals as they de-excite; receiver coils detect them.
  4. 4.A computer combines signal position and strength to construct the image.

Answer: Spatial selection by gradients, RF excitation, RF detection and computer processing produce the cross-sectional image.

Common mistakes

  • Don't say that the static field makes stationary protons begin spinning classically.
  • Don't describe the detected signal as an X-ray rather than radio-frequency emission from de-exciting protons.
  • Don't omit the gradient coils, leaving no mechanism for locating signals within the cross-section.

Exam tip

For an outline question, keep the sequence explicit: align and precess, select, excite, detect, then process.

Tier 1 · Easy

ORIGINAL

State which nuclei provide the main signal in an MR scanner and describe their behaviour in the scanner's static magnetic field.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Describe what happens to selected protons during and immediately after a short radio-frequency pulse in an MR scan.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Outline how an MR scanner obtains a cross-sectional image, beginning with a patient in the main magnetic field and ending with a computer-generated image.

[5 marks]

Total for this question: 5

3.10.5.1

The physics of diagnostic X-rays

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Thermionic emission releases electrons that accelerate through potential difference VV and decelerate at a metal target. Most energy becomes heat, so a rotating anode spreads heating.
  • Different electron energy losses produce the continuous X-ray spectrum; Emax=eVE_{\max}=eV and λmin=hc/(eV)\lambda_{\min}=hc/(eV). Characteristic lines occur when incident electrons remove inner-shell electrons and higher-shell electrons fall into vacancies.
  • Increasing tube current increases intensity and patient dose, while increasing voltage raises maximum photon energy and penetration.
  • A smaller focal spot improves sharpness but concentrates heating.
  • Filtration removes low-energy photons and collimation restricts the exposed region, controlling dose and image quality.
Diagnostic X-ray energy spectrum with a continuous distribution, characteristic lines and maximum energy.
Worked example

Calculate the maximum photon energy from a tube operating at 80kV80\,\text{kV}, in joules and electronvolts.

  1. 1.Use Emax=eV=(1.60×1019)(80×103)E_{\max}=eV=(1.60\times10^{-19})(80\times10^3).
  2. 2.This gives Emax=1.28×1014JE_{\max}=1.28\times10^{-14}\,\text{J}.
  3. 3.An electron accelerated through 80kV80\,\text{kV} gains 80keV80\,\text{keV}.

Answer: Emax=1.28×1014J=80keVE_{\max}=1.28\times10^{-14}\,\text{J}=80\,\text{keV}.

Common mistakes

  • Don't say that increasing tube current increases maximum photon energy rather than photon number.
  • Don't attribute characteristic lines to arbitrary electron deceleration instead of transitions into inner-shell vacancies.
  • Don't claim filtration removes the highest-energy photons and thereby increases skin dose.

Exam tip

In a controls question, link current to intensity, voltage to maximum energy, focal spot to sharpness and filtration or collimation to dose.

Tier 1 · Easy

ORIGINAL

State the maximum photon energy, in keV\text{keV}, from an X-ray tube operating at 68kV68\,\text{kV}.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

An X-ray tube uses an accelerating potential difference of 72kV72\,\text{kV}. Calculate the minimum X-ray wavelength. Use e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}, h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s} and c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A rotating-anode tube operates at 95kV95\,\text{kV} with a current of 3.0mA3.0\,\text{mA} for 0.080s0.080\,\text{s}. It converts 1.2%1.2\% of the electrical energy into X-rays. Calculate the maximum photon energy and the total X-ray energy produced. Explain one benefit of rotating the anode. Use e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}.

[5 marks]

Total for this question: 5

3.10.5.2

Image detection and enhancement

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In a flat-panel detector, a scintillator converts X-rays to visible photons, photodiode pixels convert light to electrical charge, and electronic scanning transfers pixel signals to a computer. The immediate digital image can be enhanced, stored and transmitted; high sensitivity and wide dynamic range may reduce repeat exposures compared with film.
  • X-rays darken photographic film, while an intensifying screen converts X-rays to visible light so less dose is needed.
  • Fluoroscopic image intensification gives live images, although prolonged viewing can raise dose.
  • X-ray-opaque barium absorbs strongly and outlines low-contrast digestive structures against soft tissue.
  • Images arise from differential transmission and absorption, not reflected X-rays.
Worked example

Explain how a flat-panel detector converts an incident X-ray pattern into a digital image.

  1. 1.The scintillator converts absorbed X-rays into visible-light photons.
  2. 2.Photodiode pixels convert the light into electrical charge proportional to local exposure.
  3. 3.Electronic scanning reads the pixel charges and a computer constructs and enhances the image.

Answer: The scintillator-photodiode-scanning chain converts spatial X-ray intensity into digital pixel values.

Common mistakes

  • Don't reverse the conversion chain and make the photodiode emit X-rays.
  • Don't say that barium improves contrast by reflecting X-rays rather than absorbing them strongly.
  • Don't claim fluoroscopy reduces dose regardless of viewing time.

Exam tip

For an operation question, give the detector stages in order and identify the energy conversion at each stage.

Tier 1 · Easy

ORIGINAL

State the function of the scintillator in a flat-panel X-ray detector.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain why a patient is given a barium suspension before an X-ray image of the digestive tract is recorded.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A flat-panel X-ray detector has a pixel pitch of 0.25mm0.25\,\text{mm}. Three adjacent photodiode rows fail, so the corresponding strip of the image records no data. Calculate the width of the missing strip and explain why a 0.50mm0.50\,\text{mm} lesion lying entirely inside it would not be recorded. Explain why image enhancement cannot reconstruct the lesion and when a repeat exposure may be justified despite the additional dose.

[5 marks]

Total for this question: 5

3.10.5.3

Absorption of X-rays

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a narrow monoenergetic beam, transmitted intensity follows I=I0eμxI=I_0e^{-\mu x}, where μ\mu is the linear attenuation coefficient and xx is thickness in compatible units. Attenuation is exponential, not a fixed subtraction per unit thickness.
  • The half-value thickness is x1/2=ln2/μx_{1/2}=\ln2/\mu; after nn half-value thicknesses, I=I0(1/2)nI=I_0(1/2)^n.
  • The mass attenuation coefficient is μm=μ/ρ\mu_m=\mu/\rho, with SI unit m2kg1\text{m}^2\,\text{kg}^{-1}, allowing materials to be compared without density's direct effect.
  • Bone and contrast media attenuate more strongly than soft tissue, so differential transmission creates image contrast.
  • The equation gives transmitted, not absorbed, intensity.
Exponential X-ray attenuation and the half-value thickness.
Worked example

An absorber has μ=0.35cm1\mu=0.35\,\text{cm}^{-1}. Calculate its half-value thickness.

  1. 1.Use x1/2=ln2/μx_{1/2}=\ln2/\mu.
  2. 2.Substitute x1/2=0.693/0.35=1.98cmx_{1/2}=0.693/0.35=1.98\,\text{cm}.

Answer: The half-value thickness is 1.98cm1.98\,\text{cm}.

Common mistakes

  • Don't subtract the same intensity for every centimetre instead of applying exponential attenuation.
  • Don't use incompatible units for μ\mu and xx.
  • Don't report I0II_0-I when the equation has calculated transmitted intensity II.

Exam tip

Check whether the question requests transmitted intensity, absorbed intensity or half-value thickness before selecting the relation.

Tier 1 · Easy

ORIGINAL

An absorber has linear attenuation coefficient 0.28cm10.28\,\text{cm}^{-1}. Calculate its half-value thickness.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A narrow X-ray beam of intensity 6.4W m26.4\,\text{W m}^{-2} passes through 7.0mm7.0\,\text{mm} of tissue with μ=0.18mm1\mu=0.18\,\text{mm}^{-1}. Calculate the transmitted intensity.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An incident X-ray beam has intensity 10.0W m210.0\,\text{W m}^{-2}. One ray crosses 18mm18\,\text{mm} of soft tissue with μ=0.050mm1\mu=0.050\,\text{mm}^{-1} and then 4.0mm4.0\,\text{mm} of bone with μ=0.32mm1\mu=0.32\,\text{mm}^{-1}. A neighbouring ray crosses 22mm22\,\text{mm} of the same soft tissue only. Calculate both transmitted intensities and the ratio of the soft-tissue-only intensity to the bone-path intensity.

[5 marks]

Total for this question: 5

3.10.5.4

CT scanner

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A CT scanner moves an X-ray tube around the patient and directs a narrow, approximately monochromatic beam through the body to an array of detectors. At many angles, detectors measure the transmitted intensity along different paths.
  • A computer combines these projections to reconstruct a cross-sectional visual image; successive slices may form a three-dimensional representation.
  • CT removes the superposition of structures in a plain radiograph and provides better localisation and useful soft-tissue information.
  • Its disadvantages are greater cost and usually a higher ionising-radiation dose than a simple X-ray image.
  • Required comparisons are limited to image resolution, cost and safety; detector construction and operation are not required.
CT acquisition using a rotating X-ray tube and detector array around a patient cross-section.
Worked example

Explain why CT can distinguish overlapping internal structures better than a plain radiograph.

  1. 1.A plain radiograph superposes attenuation from every structure along one projection.
  2. 2.CT records narrow-beam projections from many different angles.
  3. 3.Computer reconstruction assigns attenuation information to locations within a cross-sectional slice.

Answer: CT removes projection overlap by reconstructing a spatially resolved cross-section from many angular measurements.

Common mistakes

  • Don't describe CT as one wide-beam exposure from a fixed tube.
  • Don't say that CT uses non-ionising radiation.
  • Don't claim CT is always safer and cheaper than a plain radiograph despite its generally higher dose and cost.

Exam tip

A comparison must connect the reconstructed slice to improved localisation, then balance that benefit against cost and ionising dose.

Tier 1 · Easy

ORIGINAL

State why the X-ray tube moves around a patient during a CT scan.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Describe how a CT scanner produces a cross-sectional image from X-rays transmitted through a patient.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A hospital is choosing between CT and a plain X-ray image to localise a small lung lesion partly hidden by overlapping ribs, and to confirm the position of a radiopaque feeding tube. Compare the techniques using image resolution, superposition, radiation dose, cost and availability. Recommend one technique for each task.

[6 marks]

Total for this question: 6

3.10.6.1

Imaging techniques

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A tracer combines a gamma-emitting radioisotope with a compound that has affinity for a chosen organ; penetrating gamma photons leave the body for external detection. Technetium-99m emits about 140keV140\,\text{keV} gamma radiation, has a roughly 6h6\,\text{h} half-life and labels many compounds.
  • Iodine-131 is taken up by the thyroid, emits beta and gamma radiation and has a roughly 8d8\,\text{d} half-life. Indium-111 emits gamma radiation and has a roughly 2.8d2.8\,\text{d} half-life.
  • Higher gamma energy makes collimation harder because photons can penetrate the lead between adjacent collimator holes, and it lowers detection efficiency because fewer photons are fully absorbed by the detector crystal.
  • A molybdenum-technetium generator supplies fresh technetium-99m from longer-lived molybdenum-99.
  • PET detects two nearly opposite 511keV511\,\text{keV} annihilation photons in coincidence to locate positron-emitting tracer activity.
Worked example

Explain why technetium-99m is suitable for diagnostic tracer imaging.

  1. 1.Its gamma radiation is penetrating enough to leave the body for external detection.
  2. 2.Its approximately 6h6\,\text{h} half-life permits imaging but limits prolonged dose.
  3. 3.It can be attached to compounds with affinity for different organs and obtained from a hospital generator.

Answer: Detectable gamma emission, a suitably short half-life, flexible labelling and generator availability make technetium-99m useful.

Common mistakes

  • Don't choose an alpha emitter for external imaging even though alpha radiation cannot escape tissue effectively.
  • Don't say the molybdenum-technetium generator creates energy rather than supplying daughter technetium-99m from molybdenum-99 decay.
  • Don't describe PET coincidence photons as travelling in the same direction instead of nearly opposite directions.

Exam tip

Justify a tracer by linking radiation type, half-life, gamma energy and organ-specific labelling to the intended scan.

Tier 1 · Easy

ORIGINAL

State why a medical tracer used for external imaging should emit gamma radiation.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Justify the use of technetium-99m for imaging an organ. Refer to its 6h6\,\text{h} half-life, its 140keV140\,\text{keV} gamma emission and its chemical use.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Explain how a PET scan can map regions of high glucose uptake after a patient receives a positron-emitting glucose analogue. Include the nuclear event, the detection method and how position is inferred.

[5 marks]

Total for this question: 5

3.10.6.2

Half-life

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Physical half-life TPT_P is the time for half the unstable nuclei to decay and is a property of the radionuclide. Biological half-life TBT_B is the time for biological processes to remove half the tracer from an organ or body, ignoring radioactive decay.
  • Effective half-life TET_E describes the combined reduction caused by both processes.
  • The rates add, so 1/TE=1/TB+1/TP1/T_E=1/T_B+1/T_P; consequently TET_E must be shorter than both component half-lives.
  • All values must use the same time unit before reciprocals are combined.
  • The equation concerns the amount or activity remaining in the body, not a change in the radionuclide's intrinsic physical half-life.
Worked example

A tracer has TP=12hT_P=12\,\text{h} and TB=18hT_B=18\,\text{h}. Calculate TET_E.

  1. 1.Use 1/TE=1/18+1/12=5/36h11/T_E=1/18+1/12=5/36\,\text{h}^{-1}.
  2. 2.Take the reciprocal: TE=36/5=7.2hT_E=36/5=7.2\,\text{h}.
  3. 3.Check that 7.2h7.2\,\text{h} is shorter than both 12h12\,\text{h} and 18h18\,\text{h}.

Answer: The effective half-life is 7.2h7.2\,\text{h}.

Common mistakes

  • Don't add TPT_P and TBT_B directly instead of adding their reciprocals.
  • Don't use hours for one half-life and days for the other.
  • Don't obtain an effective half-life longer than a component half-life and fail to reject the result.

Exam tip

Use the shorter-than-both check immediately after evaluating an effective half-life.

Tier 1 · Easy

ORIGINAL

Define the biological half-life of a tracer in an organ.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A tracer has physical half-life 18h18\,\text{h} and biological half-life 30h30\,\text{h}. Calculate its effective half-life.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A radionuclide has physical half-life 12.0h12.0\,\text{h}. Measurements in a patient give an effective half-life of 7.20h7.20\,\text{h}. Calculate the biological half-life and the percentage of the initial activity remaining in the patient after 21.6h21.6\,\text{h}.

[5 marks]

Total for this question: 5

3.10.6.3

Gamma camera

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A gamma camera's lead collimator absorbs photons travelling in unsuitable directions, so accepted photons retain directional information; narrower holes improve spatial resolution but reduce count rate. A scintillation crystal converts each absorbed gamma photon into a visible-light flash and a light guide distributes it to photomultiplier tubes.
  • At each photocathode, light ejects photoelectrons.
  • Successive dynodes held at increasing potentials produce secondary emission, multiplying the electron number into a large anode pulse.
  • Relative photomultiplier outputs locate the flash, while pulse size estimates photon energy so scattered photons can be rejected.
  • A computer accumulates many accepted event positions to form the tracer-distribution image.
Gamma-camera chain from directional collimation to scintillation and photomultiplier detection.
Worked example

Explain the trade-off produced by making gamma-camera collimator holes narrower.

  1. 1.Narrow holes accept a smaller range of photon directions.
  2. 2.The accepted direction is better defined, improving spatial resolution.
  3. 3.Fewer photons pass through, reducing count rate and sensitivity.

Answer: Narrower holes improve spatial resolution but reduce sensitivity and require longer counting.

Common mistakes

  • Don't say the collimator focuses gamma photons like a glass lens.
  • Don't place the photocathode before the scintillation crystal and make it absorb gamma photons directly.
  • Don't claim narrower holes increase both resolution and count rate.

Exam tip

An operation answer should follow one photon through collimator, crystal, photocathode, dynodes and anode before explaining image formation.

Tier 1 · Easy

ORIGINAL

State the function of the lead collimator in a gamma camera.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Describe how a gamma photon entering a gamma camera produces an electrical pulse at the output of a photomultiplier tube.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Explain how a gamma camera determines the position of tracer activity and improves the quality of the recorded image. Include the roles of the collimator, photomultiplier array and pulse-height selection.

[5 marks]

Total for this question: 5

3.10.6.4

Use of high-energy X-rays

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • External radiotherapy directs high-energy X-rays into a tumour.
  • Their penetration delivers energy at depth, where ionisation damages DNA and prevents successful cell division.
  • Several beam directions can converge on the tumour: the target receives the combined dose, while each region of healthy tissue lies in fewer paths and receives less dose.
  • Collimation shapes each field, shielding protects selected regions, and imaging plus computer planning locate the tumour and choose beam directions.
  • Healthy tissue cannot receive zero dose, so treatment planning balances tumour control against damage to surrounding organs.
External X-ray beams from several directions converge on a tumour.
Worked example

Explain how using three beam directions reduces harm compared with delivering the entire tumour dose through one path.

  1. 1.Each beam contributes only part of the prescribed dose.
  2. 2.All three overlap at the tumour, so their doses add at the target.
  3. 3.Most surrounding tissue lies in only one beam path and receives a smaller dose.

Answer: Beam convergence maintains tumour dose while spreading the unavoidable healthy-tissue dose across different regions.

Common mistakes

  • Don't claim that surrounding healthy tissue receives no radiation.
  • Don't name collimation without explaining that it restricts the tissue inside the treatment field.
  • Don't list shielding without explaining that it limits exposure outside the planned field.

Exam tip

Use comparative dose language: full combined dose at the tumour, smaller dose in each healthy-tissue path.

Tier 1 · Easy

ORIGINAL

State why high-energy X-rays are used to treat a tumour deep inside the body.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain how directing X-ray beams at a tumour from several angles can reduce damage to healthy tissue.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A tumour of mass 0.18kg0.18\,\text{kg} receives 1.8Gy1.8\,\text{Gy} in each of 2020 treatment sessions. Calculate the total energy absorbed by the tumour. Explain two methods, other than reducing the prescribed tumour dose, that limit exposure of healthy cells. Use 1Gy=1J kg11\,\text{Gy}=1\,\text{J kg}^{-1}.

[5 marks]

Total for this question: 5

3.10.6.5

Use of radioactive implants

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Brachytherapy places sealed radioactive sources inside or very close to a tumour, producing a high local dose while reducing irradiation of distant tissue. Beta-emitting implants are suitable because beta particles are ionising but have a short range in tissue; a more penetrating gamma source would expose more healthy tissue beyond the target.
  • Several small sources can distribute dose through an irregular tumour. Activity, half-life, source position and treatment time determine the delivered dose, and temporary implants can be removed after the planned exposure.
  • Staff exposure is limited by short handling times, distance, shielding and remote handling tools.
  • The benefit must be linked to localisation of dose, not merely to beta being less penetrating.
  • When the activity falls appreciably during a treatment, the number of decays is N0(1eλt)N_0(1-e^{-\lambda t}) with N0=A0/λN_0=A_0/\lambda, not A0tA_0t.
Worked example

Explain why a beta-emitting implant may be preferred to a gamma-emitting implant for a small local tumour.

  1. 1.Beta radiation is ionising and can damage tumour-cell DNA.
  2. 2.Its short range deposits most energy close to the implanted source.
  3. 3.Gamma radiation would travel farther and irradiate more healthy tissue outside the target.

Answer: The beta implant concentrates an effective ionising dose within the local tumour while limiting distant exposure.

Common mistakes

  • Don't state only that beta is less penetrating without linking range to localised tumour dose.
  • Don't place the implant outside the patient and describe it as external-beam therapy.
  • Don't ignore source half-life and treatment time when discussing delivered dose.

Exam tip

For a justify question, connect radiation range directly to the required target size and protection of surrounding tissue.

Tier 1 · Easy

ORIGINAL

State why a beta emitter is suitable for a radioactive implant placed inside a tumour.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain why a beta-emitting implant may expose less healthy tissue than an external high-energy X-ray beam used to deliver the same tumour dose.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A temporary beta-emitting implant has initial activity 9.6MBq9.6\,\text{MBq} and physical half-life 48h48\,\text{h}. Calculate its activity after 120h120\,\text{h}. Explain how the position and radiation type of the implant help protect healthy tissue.

[5 marks]

Total for this question: 5

3.10.6.6

Imaging comparisons

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Plain X-rays are fast, available and resolve bone well, but use ionising radiation, superpose structures and give poor soft-tissue contrast. CT improves localisation through cross-sectional images but generally costs more and gives a higher ionising dose.
  • Ultrasound is portable, inexpensive, real-time and non-ionising, but is operator-dependent and transmits poorly through bone or gas.
  • MR gives excellent soft-tissue contrast without ionising radiation but is slower, costly and less convenient for some patients.
  • Radionuclide and PET scans reveal physiological function, yet require internal ionising tracers and usually have lower spatial resolution.
  • A valid comparison applies resolution, safety and convenience to the specified organ and diagnostic purpose.
Worked example

Compare ultrasound and CT for repeated imaging of a moving fetus.

  1. 1.Ultrasound is non-ionising and provides real-time images, supporting repeated monitoring.
  2. 2.CT uses ionising X-rays and generally delivers a greater dose.
  3. 3.Although CT may provide different anatomical detail, its radiation risk makes it unsuitable for routine repeated fetal imaging.

Answer: Ultrasound is preferred because it gives convenient real-time imaging without ionising dose.

Common mistakes

  • Don't declare one technique universally best without relating it to the tissue or clinical need.
  • Don't call MR an ionising X-ray technique.
  • Don't compare image quality vaguely without distinguishing spatial resolution, soft-tissue contrast and functional information.

Exam tip

Structure comparisons as paired consequences for the stated case: resolution or contrast, convenience, then ionising-radiation risk.

Tier 1 · Easy

ORIGINAL

Select a suitable imaging technique for repeated bedside monitoring of moving soft tissue when ionising radiation must be avoided.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Compare CT and MR for repeated imaging of a brain tumour. Refer to soft-tissue contrast, patient dose and convenience.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A patient may have a kidney stone. Evaluate ultrasound, CT and MR for locating the stone, using spatial resolution, soft-tissue or stone contrast, radiation dose, availability and convenience. Reach a justified recommendation.

[6 marks]

Total for this question: 6

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